Search Results

Search found 27029 results on 1082 pages for 'standard input'.

Page 48/1082 | < Previous Page | 44 45 46 47 48 49 50 51 52 53 54 55  | Next Page >

  • windows xp mode for windows 7 - save text input language settings

    - by Gero
    When I change the 'default language' in 'text services and input languages' in windows xp mode from EN-US to DE-DE the settings are reverted with the next logoff / reboot - EN-US is the default language again. Is there a way around this behaviour? I'm using the default 'XPMUser' in windows xp mode. I also checked 'turn off advanced text services' and disabled the language bar and windows xp remembers these settings - just not the default language..

    Read the article

  • Chinese IME input method in AZERTY on windows (Google IME)

    - by TimothyP
    I'm using a Belgian Azerty keyboard. The chinese input method on Mac OS works just fine, but on Windows, if I use the Google IME for example, a = z q = w etc... so it uses a qwerty layout even though my keyboard is azerty. Can I make Google IME use an azerty layout, or is there other software that uses the azerty layout instead of qwerty

    Read the article

  • Java compiler error: Can't open input server /Library/InputManagers/Inquisitor

    - by unknown (yahoo)
    I am trying to compile HelloWorld in Java under Mac OS X 10.6 (Snow Leopard) and I get this compiler error: java[51692:903] Can't open input server /Library/InputManagers/Inquisitor It happens when I am using terminal command javac and when I am trying to do this in NetBeans. I was trying to open folder "Inquisitor", but I have no access to folder, even if I login as root user. What is going on?

    Read the article

  • java compiler error: Can't open input server /Library/InputManagers/Inquisitor

    - by unknown (yahoo)
    Hi I am trying to compile helloWorld in java under snow leopard and I get this compiler error: java[51692:903] Can't open input server /Library/InputManagers/Inquisitor It happens when I am using terminal command javac and when I am trying do this in NetBeans. I was trying to open folder "Inquisitor" but I have no access to folder , even if I login as root user. Any clue what is going on? thanks.

    Read the article

  • Stream video from one app to another in OS X (Software video input device)

    - by Josh
    Is there any way to take a video stream from one application, for example VLC Media Player, and present that video stream to other applications as a video input source? For example, could I broadcast a video file from my hard disk to a website that allows video conferencing using a Flash applet? Basically, I'm looking for something like Soundflower, but for video streams. Is this possible?

    Read the article

  • Ubuntu 10.04 doesn't accept keyboard input when running under VMware on Windows 7

    - by anwar
    I have just installed Ubuntu for the first time using VMWare on Windows 7. Everything has been installed smoothly but after the installation in the login screen username is coming and when I try to enter password it is not taking any input, keyboard is not working at all. After moving away from Ubuntu keyboard and everthing else is working fine. Does anyone know what's the cause behind this ?

    Read the article

  • Set default input/output sound

    - by Nathan Koop
    I have a mac that is hooked into a sound board, I have external sound card that interfaces between the computer and the soundboard, this works great. However, whenever the computer restarts it sets the input & output to the 'internal microphone' and the 'built in output'. I'd like to default to the other soundcard. How can I do this? I'm running OS 10.6.3

    Read the article

  • Ext4 Input/Output Error Reboot via SSH

    - by LorenVS
    I've got a remote appliance, and its disk IO seems to have locked up, trying to run anything that isn't already loaded results in errors like this: $ sudo shutdown -r 0 sudo: Can't open /var/lib/sudo/<machine_name>/0: Read-only file system sudo: unable to execute /sbin/shutdown: Input/output error I have SSH access to the appliance. I'm hoping that restarting the box will fix this (if not I have to go replace the box), but trying to restart it yields the above output. Anyone have any ideas???

    Read the article

  • windows xp mode for windows 7 - save text input language settings

    - by Gero
    When I change the 'default language' in 'text services and input languages' in windows xp mode from EN-US to DE-DE the settings are reverted with the next logoff / reboot - EN-US is the default language again. Is there a way around this behaviour? I'm using the default 'XPMUser' in windows xp mode. I also checked 'turn off advanced text services' and disabled the language bar and windows xp remembers these settings - just not the default language..

    Read the article

  • Using relative path in PHP.INI doc_root -- getting No Input File Specified -- running as fastCGI

    - by J.R.
    I'm attempting to run php-cgi under LightTPD on Windows with my doc_root set to one directory up (doc_root = "../Docs") -- however, I get "No input file specified". I've set cgi.fix_pathinfo, and all the other tricks I could find with no success. If I set doc_root to an absolute path, it works fine. How can I make this work? If any additional information is required, I'll gladly provide it. Thanks in advance.

    Read the article

  • how to read input with multiple lines in java

    - by Gandalf StormCrow
    Hi all, Our professor is making us do some basic programming with java, he gaves a website and everything to register and submit our questions, for today I need to do this one example I feel like I'm on the right track but I just can't figure out the rest .. here is the actualy question : **Sample Input:** 10 12 10 14 100 200 **Sample Output:** 2 4 100 And here is what I've got so far : public class Practice { public static int calculateAnswer(String a, String b) { return (Integer.parseInt(b) - Integer.parseInt(a)); } public static void main(String[] args) { System.out.println(calculateAnswer(args[0], args[1])); } } Now I always get the answer 2 because I'm reading the single line, how can I take all lines into account? thank you For some strange reason everytime I want to execute I get this error: C:\sonic>java Practice.class 10 12 Exception in thread "main" java.lang.NoClassDefFoundError: Fact Caused by: java.lang.ClassNotFoundException: Fact.class at java.net.URLClassLoader$1.run(URLClassLoader.java:20 at java.security.AccessController.doPrivileged(Native M at java.net.URLClassLoader.findClass(URLClassLoader.jav at java.lang.ClassLoader.loadClass(ClassLoader.java:307 at sun.misc.Launcher$AppClassLoader.loadClass(Launcher. at java.lang.ClassLoader.loadClass(ClassLoader.java:248 Could not find the main class: Practice.class. Program will exit. Whosever version of answer I use I get this error, what do I do ? However if I run it in eclipse Run as Run Configuration - Program arguments 10 12 10 14 100 200 I get no output EDIT I have made some progress, at first I was getting the compilation error, then runtime error and now I get wrong answer , so can anybody help me what is wrong with this : import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.math.BigInteger; public class Practice { public static BigInteger calculateAnswer(String a, String b) { BigInteger ab = new BigInteger(a); BigInteger bc = new BigInteger(b); return bc.subtract(ab); } public static void main(String[] args) throws IOException { BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in)); String line; while ((line = stdin.readLine()) != null && line.length()!= 0) { String[] input = line.split(" "); if (input.length == 2) { System.out.println(calculateAnswer(input[0], input[1])); } } } }

    Read the article

  • Piping input to a Java app with Perl

    - by user319479
    I need to write a Perl script that pipes input into a Java program. This is related to this, but that didn't help me. My issue is that the Java app doesn't get the print statements until I close the handle. What I found online was that $| needs to be set to something greater than 0, in which case newline characters will flush the buffer. This still doesn't work. This is the script: #! /usr/bin/perl -w use strict; use File::Basename; $|=1; open(TP, "| java -jar test.jar") or die "fail"; sleep(2); print TP "this is test 1\n"; print TP "this is test 2\n"; print "tests printed, waiting 5s\n"; sleep(5); print "wait over. closing handle...\n"; close TP; print "closed.\n"; print "sleeping for 5s...\n"; sleep(5); print "script finished!\n"; exit And here is a sample Java app: import java.util.Scanner; public class test{ public static void main( String[] args ){ Scanner sc = new Scanner( System.in ); int crashcount = 0; while( true ){ try{ String input = sc.nextLine(); System.out.println( ":: INPUT: " + input ); if( "bananas".equals(input) ){ break; } } catch( Exception e ){ System.out.println( ":: EXCEPTION: " + e.toString() ); crashcount++; if( crashcount == 5 ){ System.out.println( ":: Looks like stdin is broke" ); break; } } } System.out.println( ":: IT'S OVER!" ); return; } } The Java app should respond to receiving the test prints immediately, but it doesn't until the close statement in the Perl script. What am I doing wrong? Note: the fix can only be in the Perl script. The Java app can't be changed. Also, File::Basename is there because I'm using it in the real script.

    Read the article

  • input type="text" not clickable in a jcarousel (jQuery based carousel) element

    - by AkaiKen
    Hello ! I work on a website where almost all objects are in a jCarousel element : <div> <ul id="mycarousel"> <li>object 1</li> <li>object 2</li> </ul> </div> is the "written" code, and : <div class=" jcarousel-skin-tango"> <div class="jcarousel-container jcarousel-container-horizontal" style="display: block;"> <div class="jcarousel-prev jcarousel-prev-horizontal jcarousel-prev-disabled jcarousel-prev-disabled-horizontal" style="display: block;" disabled="true"></div> <div class="jcarousel-next jcarousel-next-horizontal" style="display: block;" disabled="false"></div> <div class="jcarousel-clip jcarousel-clip-horizontal"> <ul class=" jcarousel-list jcarousel-list-horizontal" id="mycarousel" style="width: 2853px; left: 0px;"> <li class="jcarousel-item jcarousel-item-horizontal jcarousel-item-1 jcarousel-item-1-horizontal" jcarouselindex="1" style="width: 307px;"> object 1 </li> <li class="jcarousel-item jcarousel-item-horizontal jcarousel-item-1 jcarousel-item-1-horizontal" jcarouselindex="1" style="width: 307px;"> object 2 </li> </ul> </div> </div> is the generated code (here it lacks the ultimate </div>, maybe a number of rows limitation ?). I'm sorry I can't provide a link, I work on a local server currently. It works perfectly =) but... my problem is : an <input type="text" /> won't be clickable in this environment. I tried the others type of input, radio, checkbox, file : works. But text does not. I can specify a value on my input (it's its purpose on my work : a search in a database for datas to be modified and reinjected in database). In fact, I can access to the text input by clicking-right in then left. But it's pretty 'unergonomic', and in my project it's unthinkable. Ah, and I tried to style this input with a style="z-index:1000;", no changes, I think it's not a CSS problem. Does anybody have an idea ? Thank you very much in advance.

    Read the article

  • Difficulty with jQuery and input keydown event

    - by Rosarch
    I am making a simple JavaScript enhanced list. I want it to be a list of inputs, each with an 'Add' and 'Remove' button. If the user clicks 'Add', a new li will be added. If the user clicks 'Remove', that li will be removed. It works fine, except for hitting "enter" in an <input>. Currently, it always causes the Remove.click event handler to fire, unless there's only one item in the list. I'm not sure why. How can I suppress this? Here is the complete jQuery. My attempt to fix the "enter" issue is commented out, and it doesn't work. I suspect that I could be designing this code better; if you see an improvement I'd love to hear it. function make_smart_list(list) { var ADD_CLASS = 'foo-widget-Add'; var REMOVE_CLASS = 'foo-widget-Remove'; var jq_list = $(list); jq_list.parents('form').submit(function() { return false; }); function refresh_handlers() { jq_list.find(sprintf('.%s, .%s', REMOVE_CLASS, ADD_CLASS)).unbind('click'); // jq_list.find('input').unbind('submit'); // // jq_list.find('input').submit(function() { // var jq_this = $(this); // var next_button = jq_this.nextAll('button'); // if (next_button.hasClass(ADD_CLASS)) { // next_button.nextAll('button').click(); // return; // } // // if (next_button.hasClass(REMOVE_CLASS)) { // return false; // } // // }); jq_list.find("." + REMOVE_CLASS).click(function() { var jq_this = $(this); jq_this.parent().remove(); refresh_handlers(); return false; }); jq_list.find("." + ADD_CLASS).click(function() { var jq_this = $(this); if (jq_this.prevAll('input').val() == '') { return; } jq_this.parent().clone().appendTo(jq_this.parent().parent()); jq_this.parent().next().find('input').val('').focus(); jq_this.removeClass(ADD_CLASS).addClass(REMOVE_CLASS); jq_this.text('Remove'); refresh_handlers(); return false; }); } refresh_handlers(); } (sprintf is another script I have.)

    Read the article

  • How to implement a UINavigationController to a standard UIView

    - by tarnfeld
    This is the structure of my application currently: UIWindow UIViewController (Root View Controller) UINavigationController UITableView UIViewController (PresentModalViewControllerAnimated:YES) UITableView This is how I want it to be: UIWindow UIViewController (Root View Controller) UINavigationController UITableView UIViewController (PresentModalViewControllerAnimated:YES) UINavigationController UITableView I have a view that slides up and I want that view to have its own UINavigationController. It's for the app settings so I want to have nested options. Any ideas how to do this? The application type was a Navigation app to start with which is where the Root View Controller's UINavigationController came from.

    Read the article

  • Reading non-standard elements in a SyndicationItem with SyndicationFeed

    - by Jared
    With .net 3.5, there is a SyndicationFeed that will load in a RSS feed and allow you to run LINQ on it. Here is an example of the RSS that I am loading: <rss version="2.0" xmlns:media="http://search.yahoo.com/mrss/"> <channel> <title>Title of RSS feed</title> <link>http://www.google.com</link> <description>Details about the feed</description> <pubDate>Mon, 24 Nov 08 21:44:21 -0500</pubDate> <language>en</language> <item> <title>Article 1</title> <description><![CDATA[How to use StackOverflow.com]]></description> <link>http://youtube.com/?v=y6_-cLWwEU0</link> <media:player url="http://youtube.com/?v=y6_-cLWwEU0" /> <media:thumbnail url="http://img.youtube.com/vi/y6_-cLWwEU0/default.jpg" width="120" height="90" /> <media:title>Jared on StackOverflow</media:title> <media:category label="Tags">tag1, tag2</media:category> <media:credit>Jared</media:credit> <enclosure url="http://youtube.com/v/y6_-cLWwEU0.swf" length="233" type="application/x-shockwave-flash"/> </item> </channel> When I loop through the items, I can get back the title and the link through the public properties of SyndicationItem. I can't seem to figure out how to get the attributes of the enclosure tag, or the values of the media tags. I tried using SyndicationItem.ElementExtensions.ReadElementExtensions<string>("player", "http://search.yahoo.com/mrss/") Any help with either of these?

    Read the article

  • Building a blog: what's standard?

    - by Charlotte
    I'm generally pretty new at web stuff. I wanted to build a blog from scratch to get some practice. Few questions: Do most people add new entries of a blog by directly editing the html or is there a more dynamic way of doing this that is used more frequently? I'm assuming you can store the entries in some type of database and then display them via javascript or something similar? What are the most frequently used tools for what I'm describing? I know its about as simple as it gets, but like most things, I just need some tips to get started. Thanks!

    Read the article

  • Guru Of the Week n° 41 : utiliser la bibliothèque standard, un article de Herb Sutter traduit par la rédaction C++

    La bibliothèque standard fournit un nombre important de structures de données et d'algorithmes. Dans de nombreux cas, il est possible de remplacer les structures de contrôle du langage (if, for, while) par les fonctionnalités provenant de celle-ci. Dans ce Guru Of the Week n° 41, Herb Sutter lance le défi de créer un Mastermind en minimisant l'utilisation des structures de contrôle. Guru Of the Week n° 41 : utiliser la bibliothèque standard Saurez-vous relever le défi et proposer un tel code de Mastermind ? Retrouver l'ensemble des Guru of the Week sur la

    Read the article

  • How to detect touches on MPMoviePlayerController window and still having the standard playback cont

    - by gkedmi
    Hello I have the MPMoviePlayerController set up to play a movie.I want to detect a touch on the movie for bringing up few buttons.I used the code : // The movie's window is the one that is active UIWindow* moviePlayerWindow = [[UIApplication sharedApplication] keyWindow]; // Now we create an invisible control with the same size as the window UIControl* overlay = [[[UIControl alloc] initWithFrame:moviePlayerWindow.frame]autorelease]; // We want to get notified whenever the overlay control is touched [overlay addTarget:self action:@selector(movieWindowTouched:) forControlEvents:UIControlEventTouchDown]; // Add the overlay to the window's subviews [moviePlayerWindow addSubview:overlay]; but then the play back controllers doesn't appear , I guess because the player window doesn't get the touch.how can I keep the player controllers and still detects touches? thanks

    Read the article

< Previous Page | 44 45 46 47 48 49 50 51 52 53 54 55  | Next Page >